How do you solve Perfect Squares on LeetCode?

Perfect Squares (LeetCode #279) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n√n) time complexity and O(n) space complexity. Key insight: Understanding perfect squares is crucial for solving this problem efficiently.

What is the brute force solution for Perfect Squares?

The brute force approach for Perfect Squares is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to find all combinations of perfect squares that sum up to n. This method is straightforward but inefficient as it explores many unnecessary paths. The optimal Optimal Solution approach improves this to O(n√n).

What algorithm or data structure is used to solve Perfect Squares?

Perfect Squares uses the following concepts: Math, Dynamic Programming, Breadth-First Search. The recommended patterns to study are: Dynamic Programming, Backtracking.

What are the interview tips for Perfect Squares?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as the smallest values of n. Explain your thought process as you code; it shows your understanding and helps the interviewer follow along.

What are common mistakes when solving Perfect Squares?

Not considering all perfect squares when calculating combinations. Failing to initialize the dp array correctly, leading to incorrect results.

#279

Perfect Squares

Medium

MathDynamic ProgrammingBreadth-First SearchDynamic ProgrammingBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n√n)
Space	O(1)	O(n)

💡

Intuition

Time O(n√n)Space O(n)

The optimal solution uses dynamic programming to build up the solution from smaller subproblems. It efficiently calculates the minimum number of perfect squares needed for each number up to n.

⚙️

Algorithm

3 steps

1Step 1: Create an array dp where dp[i] represents the least number of perfect squares that sum to i.
2Step 2: Initialize dp[0] = 0 and fill dp[i] for all i from 1 to n using previously computed values.
3Step 3: For each i, iterate through all perfect squares less than or equal to i and update dp[i] accordingly.

solution.py10 lines

1# Full working Python code
2from math import isqrt
3
4def numSquares(n):
5    dp = [float('inf')] * (n + 1)
6    dp[0] = 0
7    for i in range(1, n + 1):
8        for j in range(1, isqrt(i) + 1):
9            dp[i] = min(dp[i], dp[i - j * j] + 1)
10    return dp[n]

ℹ

Complexity note: The time complexity is O(n√n) because for each number up to n, we check all perfect squares up to that number, leading to a linear scan for each square.

1Understanding perfect squares is crucial for solving this problem efficiently.
2Dynamic programming allows us to build solutions incrementally, reducing redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.